A classical mechanics problem by Soon Yu

Classical Mechanics Level 3

A 40kg slab rests on a frictionless floor. A 10kg block rests on top of it. The coefficient of static friction between block and slab is 0.60, whereas their kinectic friction of coefficient is 0.40. If the 10kg block is pulled to left by a horizontal force of 100N, what is the acceleration of the slab?


The answer is 1.

Soon Yu by Soon Yu , 28, Malaysia

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2 solutions

assume g = 10 m / s 2 g = 10 m/s^{2} then F s = 10 × 10 × 0.6 = 60 F_{s} = 10 \times 10 \times 0.6 = 60 F s < F F_{s} < F

so the block will move with kinetic frictional force working on it as much as F k = 10 × 10 × 0.4 = 40 F_{k} = 10 \times 10 \times 0.4 = 40

The frictional force will also works on the slab, then F = ( m s l a b ) a F = (m_{slab}) a 40 = 40 a 40 = 40 a a = 1 a = 1

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When 100 N 100N force is applied to the 10 k g 10kg block it overcomes the static frictional force which is μ s m g = 0.6 × 10 × 10 = 60 N \mu_s m g = 0.6\times 10 \times 10 = 60N (assuming g = 10 m s − 2 g = 10 ms^{-2} ). Therefore, the 10 k g 10kg block moves to the left and the kinetic frictional force of μ k m g = 0.40 × 10 × 10 = 40 N \mu_k mg = 0.40\times 10 \times 10 = 40N acts on the slab accelerating it by 40 M = 40 40 = 1 m s − 2 \dfrac {40} {M} = \dfrac {40}{40} = \boxed{1} ms^{-2}

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